This was my favorite of all the ShmooCon IX puzzles. It reminds me of the logic grid puzzles I used to do in elementary school. Bobby has brown hair. The person who likes green has blue eyes, etc.

## Directions

The numbers next to each row and column indicate the size of groups of contiguous black pixels. Fill in the pixels.

For example: "2-1" in a five-square row could indicate any of three possible arrangements:

We're given a 21x21 grid of squares. The content of each row is described on the right side of the grid, and the content of each column is described at the bottom. Each number in the row/column description describes exactly that many black squares, while each "-" mark describes one or more white squares.

The number of total squares in any one row or column is exactly 21, so the sum of the count of black squares for any given row or column plus the sum of the "-" marks for that row or column must not be greater than 21. Furthermore, any row or column description whose sum of the count of black squares and "-" marks equals exactly 21 has been described to us in the final state. So let's look for any rows or columns that add up to 21 and go ahead and fill them.

As shown above, `ROWS 3` and `7` and `COLUMNS E` and `G` were
described in their exact state so all we had to do was fill them in as
described. So now that we've exhausted our easy rows, let's move on to
the next easiest case. Look for any rows or columns that add to 20, and
draw the possible outcomes. If any squares overlap in all possible
states for a given row or column, then we know that square must be
black. Squares that exist in some but not all possibilities may or may
not be black squares. We'll mark them with blue to keep them in mind.

`ROW 1`'s description adds to 20. All possible solutions to `ROW 1`
are presented above. We'll mark any square that exists in all possible
solutions in black, and we'll keep the remaining possibilities in blue.
Now we'll transfer the result back to the overall puzzle and check our
progress.

`ROW 19` also adds to 20. We'll perform the same possibility check on
it.

Now transfer it back to the overall puzzle.

That takes care of all the rows that add to 20. Now let's search for columns.

`COLUMN D` fits the bill. Once again, we perform possibility analysis
and transfer the result to the overall puzzle.

We've now exhausted the rows and columns that add to 20. Let's keep going down this route and look for rows or columns that add to 19.

`ROWS 4` and `5` are identical, so we'll take the solution from
`ROW 4` and apply it to `ROWS 4` and `5` in the overall puzzle.

`ROW 15` also adds to 19. Let's analyze possible solutions.

And again, add the result back into the overall puzzle.

We've now reached the point of tedium. Progressing along the brute-force path of mapping out possible outcomes now requires too much effort. It's time to find an alternate way to attack the puzzle.

Instead of examining an entire row or column at a time, let's look at
portions of a row or column now that we've filled out some of the
puzzle. Take `COLUMN A`, for example. We know the first contiguous
group of black squares consists of 7 squares. We also know that squares
`A3`, `A4`, `A5`, and `A7` are black. Based on these two pieces
of information, we can conclude that the first contiguous square of
`COLUMN A` starts at `A1` and ends at `A7`, starts at `A3` and
ends at `A9`, or something in between. Regardless, square `A6` must
be black.

Moving along, let's examine `COLUMN D`. We've already satisfied the
`1-3-1` conditions at beginning and end, so all we need is a group of
4 black squares in the middle. Since `D15` is a single square,
`D14` cannot be black and therefore must be white.

Let's suppose that square `I1` is black. If so, that means that
square `I3` is the start of a group of 4 black squares. However, if
we fill in squares `I4`, `I5`, and `I6`, we'll end up with a
group of 5 black squares from `I3` to `I7`. This violates the
description of `COLUMN I`, so we have proven that `I1` cannot be
black and therefore must be white. `I3` must be the first single
black square in `ROW I`, and `I7` is part of a 4-square group.
Since `I3` is a single black square, `I4` cannot be black and
therefore must be white.

`COLUMN L` starts with 4 black squares. Since `L1` is black, the
next three squares (`L2`, `L3`, and `L4`) must also be black.
`L5` cannot be black and therefore must be white.

`COLUMN O` starts with 7 black squares. Since `O1` is black, the
next six squares (`O2`, `O3`, `O4`, `O5`, `O6`, and `O7`)
must also be black.

`COLUMNS R` and `S` start with a single black square, and the next
group consists of 3 black squares. Since `R3` and `S3` are black,
the subsequent two squares (`R4`, `R5` and `S4`, `S5`,
respectively) must be black.

`COLUMN U` contains an initial group of 7 black squares. We can
analyze it the same as we analyzed `COLUMN A`, and we conclude that
`U4`, `U5`, and `U6` must all be black. These changes are
depicted in the next image.

There isn't enough room between `L1` and `O1` for a group of 2
black squares plus white padding, so `L1` must be part of a group of
2 black squares. Since `I1` is white, `J1` must be a single black
square. Therefore `K1` must be white and `M1` finishes out the
group of 2 with `L1`. `N1` must then be white, and `U1` must then
in turn be black.

In `ROW 4`, `L4` is a single black square, so `K4` and `M4`
must be white. `J4` must be the final single black square in
`ROW 4`. Because the `1-3-1` at the end of `ROWS 4` and `5` is
satisfied, `P4`, `T4` and `P5`, `T5` must be white.

To finish out the initial group of 4 black squares in `COLUMN J`,
`J2` must be black, and `J5` must be white.

`U2` must be black to finish out the initial group of 7 black squares
in `COLUMN U`.

By now, the pattern here should look pretty familiar. It's quite clearly a QR code. Maybe we can get some help if we do an internet search for 21x21 QR codes!

I found some helpful information at QRStuff.com. Specifically, I found the image below that describes QR code data standards.

Based on the QR code depiction, we can go ahead and fill in our top left, top right, and bottom left corners.

`K2` must be black to complete `ROW 2`.

`COLUMN C` has a group of 2 black squares somewhere between `C9`
and `C13`. `C12` cannot be black because the first group in
`ROW 12` contains only 22 black squares, and that condition has
already been satisfied by `D12` and `E12`. `C13` cannot be black,
because `C12` above and `C14` below must be white. `C11` cannot
be black, because the group of 5 black squares in `ROW 11` cannot
start that close to the left edge, because there isn't sufficient room
for previous black squares and white space. Therefore `C9` and
`C10` must be black to complete `COLUMN C`.

`COLUMN F` also has a group of 2 black squares somewhere between
`F9` and `F13`. `F11`, `F12`, and `F13` must be white to
satisfty conditions on `ROWS 11`, `12`, and `13`, respectively.
Therefore `F9` and `F10` must be black to complete `COLUMN F`.

`A9` and `B9` must be black to satisfy the initial group of 3 black
squares in `ROW 9`. This completes `COLUMN A`.

Now that we've satisfied the initial group of 3 black square in
`ROW 9`, we know that `E9` through `G9` are part of a group of 8
black squares. `H9` through `L9` must also be black to complete
that group of 8.

Because `COLUMN G` is completed, `G10` cannot be used to satisfy
the initial group of 5 black squares in `ROW 10`. Therefore `B10`
must be black. Now a single black square remains in `COLUMN B`,
somewhere between `B12` and `B13`. Again, as the first condition in
`ROW 12` has already been met, `B12` must be white. Therefore
`B13` must be black to complete `COLUMN B`.

With the first two conditions of `ROW 9` satisfied, `D9` must be
white. Therefore `D13` must be black to complete `COLUMN D`.

So we've now finished the left third of puzzle and most of the top
third. The rest is looking pretty sparse though. It looks like I missed
another easy one: `ROW 8` has two groups of 2 black squares. Based on
the QR code depiction above, `A8` through `H8` and `N8` through
`U8` must be white. That means that the two groups of 2 must exist
between `I8` and `M8`. That only works out if `I8`, `J8` and
`L8`, `M8` are black, leaving `K8` white in the middle to
complete `ROW 8`.

Knowing that `L8` and `L9` are black, we can deduce that `L10`
must be black to satisfy the group of 3 black squares in `COLUMN L`.

We now know that `I5` cannot be black without violating the
description of `COLUMN I`, so `I5` must be white. Therefore `K5`
and `M5` must be black to complete `ROW 5`.

Because `M5` is black, `M6` must also be black to satisfy a group
of 4 black squares in `COLUMN M`. This completes `ROW 6`. With
`ROW 6` completed, `I10` must be black to satisfy a group of 4
black squares in `COLUMN I`.

Now that we've made some decent progress, let's go back and check out
some possibilities. `COLUMN K` looks like a good place to start.

And transfer it back to the overall puzzle.

Now that we've satisfied the second group of 3 black squares in
`ROW 19`, we know that `I19` must be a single black square, `J19`
must be white, and `N19` must be white.

Let's take a closer look at `ROW 11`. The last three groups consist
of `2-5-2`. There's not enough room for those three sets plus white
squares to exist between `M11` and `U11`, so some of the black
squares must occur to the left of `M11`. `I11` and `L11` cannot
be black without violating descriptions of `COLUMNS I` and `L`,
respectively, so we can conclude that `ROW 11` must contain a group
of 2 black squares from `J11` to `K11`. This satisfies a group of 4
black squares for `COLUMN K`, so `K15` must be white.

Now let's look at `COLUMN L`. `L11`, `L12`, and `L13` cannot be
black and therefore must be white. That leaves exactly enough room to
use `L14` through `L21` to complete `COLUMN L`.

`ROW 21` requires that `L21` be a single black square, so `K21`
must be white. Therefore `K18` must be black to satisfy the group of
3 in `COLUMN K`. `K17` must then be white and `K16` must be black
to complete `COLUMN K`.

The description of `ROW 14` requires a single black square somewhere
to the left of `J14`. Based on the QR code depiction above, `A14`
through `H14` must be white. `I14` must be the initial black square
in `ROW 14`, and `M14` must be black to satisfy the group of 3.

`COLUMN J` requires a group of 3 black squares at the end of the
column. Since `J14` and `J19` cannot be black, that group of 3 has
to exist somewhere between `J15` and `J18`. We can conclude that
`J16` and `J17` must be black, and the group is finished with
either `J15` or `J18`.

Now that we've determined `J17` is black, we know that it must be
part of a group of 2 black squares. `I17` must also be black to
finish this group.

Back to possibility analysis. Let's check out `ROW 11`.

And transfer it back to the overall puzzle.

Back to `ROW 10`. We need to place a group of 2 and a group of 3
black squares somewhere between `N10` and `U10`. `N10` and
`O10` cannot be black without violating the descriptions of
`COLUMNS N` and `O`, respectively, so they must be white. That
leaves exactly enough room to complete `ROW 10` using the squares
from `P10` to `U10`.

`I14` is part of a group of 4 black squares in `COLUMN I`. However,
`I14` cannot be the last in the group, since `I11` must be white.
Therefore `I15` must also be black. This means that `J15` must be
white, and `J18` must be black to complete `COLUMN J`.

`M19` is part of a group of 3 black squares in `COLUMN M`. However,
`M19` cannot be the first in the group, since `M21` must be white.
Therefore `M18` must also be black. This satisfies the group of 4
black squares in `ROW 18`, so `I18` must be white. `I16` and
`I20` through `I21` must be black to complete `COLUMN I`.

Based on the QR depiction above, `H16` must be white. Now that we
know `I16` is black, `M16` must also be black and `ROW 16` is
complete.

Since `M17` must be white to separate the groups of 4 and 3 black
squares in `COLUMN M`, we know that `M20` must be black to finish
the group of 3, and `M15` and `M13` must be black to finish the
group of 4. `M10` cannot be black, so `M11` must be black to
complete `COLUMN M`.

With the group of 5 black squares finished in `ROW 11`, `R11` must
be white.

Since `M15` is black, `N15` must be white. That leaves exactly
enough space from `Q15` to `U15` to complete `ROW 15`.

`Q10` through `Q11` and `Q15` are all part of the same group of 6
black squares in `COLUMN Q`. We can fill in `Q12` through `Q14`
to finish out that group.

`P15` is part of a group of 3 black squares in `COLUMN P`. `P16`
must be white, so `P13` and `P14` must be black to finish the group
of 3. There isn't enough room between `P15` and `P19` for a group
of 2 black squares with white separators, so `P19` must be part of a
group of 2. In order to have room for the single black square at the
end of `COLUMN P`, `P20` must be white. `P18` must be black to
finish off the group of 2, and `P21` is black to complete
`COLUMN P`.

`U10` is part of a group of 4 black squares in `COLUMN U`. However,
`U10` cannot be the last in the group, since `U8` must be white.
`U11` and `U12` must also be black. This completes `ROW 11`.

Since `S11` must be white, `S9` must be black to satisfy the group
of 2 black squares in `COLUMN S`.

To complete `ROW 14`, one of the squares between `S14` and `T14`
must be black. `T14` cannot be black because `T12` through `T13`
don't provide enough space for a group of 1 black square with white
space on either side. Therefore `S14` is black and this completes
`ROW 14`.

`S14` also completes `COLUMN S`, so `S19` must be white. `O19`
must be black to satisfy the group of 4 black squares in `ROW 19`.

With `O19` black, `O18` must be white. Therefore `Q18` must be
black to satisfy the group of 2 black squares in `ROW 18`.

`Q21` cannot be black without violating the description of
`COLUMN Q`. `O21` cannot be black because there is not enough room
to squeeze another single black square to the left of it. Therefore
`P21` is a single black square and a group of 2 black squares must
exist somewhere between `R21` and `U21`. However, `S21` cannot be
black because `COLUMN S` is completed. `T21` and `U21` must be
black to complete `ROW 21`.

With `T21` black, `T20` must also be black to satisfy the final
group of 2 black squares in `COLUMN T`. This completes `ROW 20`.
`T19` must then be white, so `U19` must be black to complete
`ROW 19`.

With `ROW 20` completed, `Q17` must be black to complete
`COLUMN Q`.

To complete `ROW 18`, a single black square must exist somewhere
between `S18` and `T18`. However, `S18` can't be black because
`COLUMN S` is completed. `T18` must be black to complete
`ROW 18`.

With `T18` black, `T17` must be black to complete `COLUMN T`.

With `T17` black, `Q17` must be part of a group of 2 black squares.
Because `COLUMN P` is completed, `R17` must be black to complete
`ROW 17`. This also completes `COLUMN R`.

With `COLUMNS R` and `T` completed, `S9` must be a single black
square in `ROW 9`. To complete `ROW 9`, a group of 2 black squares
must exist somewhere between `N9` and `Q9`. The only two contiguous
squares in uncompleted columns are `N9` and `O9`, which must both
be black to complete `ROW 9`. This also completes `COLUMNS N` and
`O`.

With `ROW 9` completed, `U13` must be black to complete
`COLUMN U` and `ROW 13`.

Finally, `H12` must be black to complete `ROW 12` and `COLUMN H`.

And here's the final solution, without the clutter.